Karatsuba Multiplication Algorithm – Python Code

Motivation for this blog post

I’ve enrolled in Stanford Professor Tim Roughgarden’s Coursera MOOC on the design and analysis of algorithms, and while he covers the theory and intuition behind the algorithms in a surprising amount of detail, we’re left to implement them in a programming language of our choice.

And I’m ging to post Python code for all the algorithms covered during the course!

The Karatsuba Multiplication Algorithm

Karatsuba’s algorithm reduces the multiplication of two n-digit numbers to at most  n^{\log_23}\approx n^{1.585} single-digit multiplications in general (and exactly n^{\log_23} when n is a power of 2). Although the familiar grade school algorithm for multiplying numbers is how we work through multiplication in our day-to-day lives, it’s slower (\Theta(n^2)\,\!) in comparison, but only on a computer, of course!

Here’s how the grade school algorithm looks:
(The following slides have been taken from Tim Roughgarden’s notes. They serve as a good illustration. I hope he doesn’t mind my sharing them.)


…and this is how Karatsuba Multiplication works on the same problem:



A More General Treatment

Let x and y be represented as n-digit strings in some base B. For any positive integer m less than n, one can write the two given numbers as

x = x_1B^m + x_0
y = y_1B^m + y_0,

where x_0 and y_0 are less than B^m. The product is then

xy = (x_1B^m + x_0)(y_1B^m + y_0)
xy = z_2B^{2m} + z_1B^m + z_0


z_2 = x_1y_1
z_1 = x_1y_0 + x_0y_1
z_0 = x_0y_0

These formulae require four multiplications, and were known to Charles Babbage. Karatsuba observed that xy can be computed in only three multiplications, at the cost of a few extra additions. With z_0 and z_2 as before we can calculate

z_1 = (x_1 + x_0)(y_1 + y_0) - z_2 - z_0

which holds since

z_1 = x_1y_0 + x_0y_1
z_1 = (x_1 + x_0)(y_1 + y_0) - x_1y_1 - x_0y_0

A more efficient implementation of Karatsuba multiplication can be set as xy = (b^2 + b)x_1y_1 - b(x_1 - x_0)(y_1 - y_0) + (b + 1)x_0y_0, where b = B^m.


To compute the product of 12345 and 6789, choose B = 10 and m = 3. Then we decompose the input operands using the resulting base (Bm = 1000), as:

12345 = 12 · 1000 + 345
6789 = 6 · 1000 + 789

Only three multiplications, which operate on smaller integers, are used to compute three partial results:

z2 = 12 × 6 = 72
z0 = 345 × 789 = 272205
z1 = (12 + 345) × (6 + 789) − z2z0 = 357 × 795 − 72 − 272205 = 283815 − 72 − 272205 = 11538

We get the result by just adding these three partial results, shifted accordingly (and then taking carries into account by decomposing these three inputs in base 1000 like for the input operands):

result = z2 · B2m + z1 · Bm + z0, i.e.
result = 72 · 10002 + 11538 · 1000 + 272205 = 83810205.

Pseudocode and Python code


Teach Yourself Machine Learning the Hard Way!

This formula is kick-ass!

Darshan Hegde

It has been 3 years since I have steered my interests towards Machine Learning. I had just graduated from college with a Bachelor of Engineering in Electronics and Communication Engineering. Which is, other way of saying that I was:

  • a toddler in programming.
  • little / no knowledge of algorithms.
  • studied engineering math, but it was rusty.
  • no knowledge of modern optimization.
  • zero knowledge of statistical inference.

I think, most of it is true for many engineering graduates (especially, in India !). Unless, you studied mathematics and computing for undergrad.

Lucky for me, I had a great mentor and lot of online materials on these topics. This post will list many such materials I found useful, while I was learning it the hard way !

All the courses that I’m listing below have homework assignments. Make sure you work through each one of them.

1. Learn Python

If you are new to programming…

View original post 507 more words

Magic 5-gon Ring — Project Euler (Problem 68)

Yet another exciting math problem that requires an algorithmic approach to arrive at a quick solution! There is a pen-paper approach to it too, but this post assumes we’re more interested in discussing the programming angle.

First, the problem:

Working clockwise, and starting from the group of three with the numerically lowest external node (4,3,2 in this example), each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.

It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.

Total Solution Set:
9 4,2,3; 5,3,1; 6,1,2
9 4,3,2; 6,2,1; 5,1,3
10 2,3,5; 4,5,1; 6,1,3
10 2,5,3; 6,3,1; 4,1,5
11 1,4,6; 3,6,2; 5,2,4
11 1,6,4; 5,4,2; 3,2,6
12 1,5,6; 2,6,4; 3,4,5
12 1,6,5; 3,5,4; 2,4,6
By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.


Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum 16-digit string for a “magic5-gon ring?


In attempting this problem, I choose to label the 5 inner nodes as i, j, k, l, and m.
α, β, γ, δ, and θ being the corresponding outer nodes.

Let x be the sum total of each triplet line, i.e.,

x = α + i + j = β + j + k = γ + k + l = δ + l + m = θ + m + i


First Observation:
For the string to be 16-digits, 10 has to be in the outer ring, as each number in the inner ring is included in the string twice. Next, we fill the inner ring in an iterative manner.

Second Observation:
There 9 numbers to choose from for the inner ring — 1, 2, 3, 4, 5, 6, 7, 8 and 9.
5 have to be chosen. This can be done in 9C5 = 126 ways.
According to circular permutation, if there are n distinct numbers to be arranged in a circle, this can be done in (n-1)! ways, where (n-1)! = (n-1).(n-2).(n-3)…3.2.1. So 5 distinct numbers can be arranged in 4! permutations, i.e., in 24 ways around a circle, or pentagonal ring, to be more precise.
So in all, this problem can be solved in 126×24 = 3024 iterations.

Third Observation:
For every possible permutation of an inner-ring arrangement, there can be one or more values of x (triplet line-sum) that serve as a possible contenders for a “magic” string whose triplets add up to the same number, x. To ensure this, we only need that the values of α through θ of the outer ring are distinct, different from the inner ring, with the greatest of these equal to 10.
Depending on the relative positioning of the numbers in the inner ring, one can narrow the range of x-values one might have to check for each permutation. To zero-down on such a range, let’s look at an example. Shown in the figure below is a randomly chosen permutation of number in the inner ring – 7, 2, 3, 4 and 5, in that order.


So 10, 9, 8, 6 and 1 must fill the outer circle. It’s easy to notice that the 5, 7 pair is the greatest adjacent pair. So whatever x is, it has to be at least 5 + 7 + 1 = 13 (1 being the smallest number of the outer ring). Likewise,  2, 3 is the smallest adjacent pair, so whatever x is, it can’t be any more than 2 + 3+ 10 = 15 (10 being the largest number of the outer ring). This leaves us with a narrow range of x-values to check – 13, 14 and 15.

Next, we arrange the 5 triplets in clock-wise direction starting with the triplet with the smallest number in the outer ring to form a candidate string. This exercise when done for each of the 3024 permutations will shortlist a range of candidates, of which, the maximum is chosen.

That’s all there is to the problem!

Here’s the Python Code. It executes in about a tenth of a second!

Ans: 6531031914842725

Maximum Path Sum — Dynamic Programming Algorithm

I came across this problem recently that required solving for the maximum-sum path in a triangle array.

To copy the above triangle array:

As can be seen, there are 15 levels to this tree (including the top most node). Therefore, there are 214 possible routes to scan in order to check for the maximum sum using brute force. As there are only 214 (16384) routes, it is possible to solve this problem by trying every route. However, doing the same using brute force on a triangle array of 100 levels would take several billion years to solve using a computer that checks through say, 1012 routes per second. A greedy algorithm might per-chance work for the particular 4-level example problem stated above, but will not always work, and in most cases won’t. For instance, for the 100-level problem:

The Algorithm

Solving such a problem would require a powerful approach – and surely enough, there is an algorithm that solves the 100-level problem in a fraction of a second. Here’s a brief sketch of the algorithm:

You have such triangle:

  7 4
 2 4 6  
8 5 9 3

Let’s say you’re on the penultimate level 2 4 6 and you have to iterate over it.

From 2, you can go to either 8 or 5, so 8 is better (maximize you result by 3) so you calculate the first sum 8 + 2 = 10

From 4, you can go to either 5 or 9, so 9 is better (maximize you result by 4) so you calculate the second sum 9 + 4 = 13

From 6, you can go to either 9 or 3, so 9 is better again (maximize you result by 6) so you calculate the third sum 9 + 6 = 15

This is the end of first iteration and you got the line of sums 10 13 15.

Now you’ve got triangle of lower dimension:

   7    4
10   13    15  

Keep going this way…

     20    19

…and you finally arrive at 23 as the answer.

The Code

Now for the Python code. I first store the 100-level triangle array in a text file, euler67.txt
I read the triangle array into Python and successively update the penultimate row and delete the last row according to the algorithm discussed above.

This code is the key to solving problems 18 and 67 of Project Euler.
Problem 18
Ans: 1074
Problem 67
Ans: 7273

Collatz Conjecture — What You Need to Know

Like many of my previous posts, this post too has something to do with a Project Euler problem. Here’s a sketch of the Colatz Conjecture.

The following iterative sequence is defined for the set of positive integers:

n → n/2 (n is even)
n → 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

So basically, it’s just this. Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has aptly been called oneness! But perhaps oneness has its pitfalls too…

If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

It can be seen that the sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.


Histogram of stopping times for the numbers 1 to 100 million. Stopping time is on the x axis, frequency on the y axis.

Approach 1 (A naïve, but straigh forward method)

Approach 2 (Smart, quick method that uses dynamic programming with the help of dictionaries)

I couldn’t help appreciate the elegance of the second algorithm. It’ll be well worth perusing if you don’t get it at one go. [Hint: It keeps track of the number of terms of a particular sequence as values assigned to keys of a Python dictionary]

Ans: 837799

Highly Divisible Triangular Number — Project Euler (Problem 12)

All n numbers are Triangle Numbers. They’re called so, because they can be represented in the form of a triangular grid of points where the first row contains a single element and each subsequent row contains one more element than the previous one.


Problem 12 of Project Euler asks for the first triangle number with more than 500 divisors.

These are the factors of the first seven triangle numbers:

1 = 1: 1
2 = 3: 1,3
3 = 6: 1,2,3,6
4 = 10: 1,2,5,10
∑5 = 15: 1,3,5,15
∑6 = 21: 1,3,7,21
∑7 = 28: 1,2,4,7,14,28

Here’s how I proceeded:

First Step: Find the smallest number with 500 divisors. Seems like a good starting point to begin our search.
Second Step: Starting at the number found in the previous step, search for the next triangle number. Check to see whether this number has 500+ divisors. If yes, this is the number we were looking for, else…
Third Step: Check n for which ∑n = triangle number found in the previous step
Fourth Step: Add (n+1) to the last triangle number found, to find the next triangle number. Check whether this number has 500+ divisors. If yes, this number is the answer. If not, repeat Fourth Step till the process terminates.

Now for the details:

The First Step isn’t exactly a piece of cake, but necessary to reduce computation time. I solved this with a bit of mental math. The main tool for the feat is the prime number decomposition theorem:

Every integer N is the product of powers of prime numbers

N = pαqβ· … · rγ
Where p, q, …, r are prime, while α, β, …, γ are positive integers. Such representation is unique up to the order of the prime factors.
If N is a power of a prime, N = pα, then it has α + 1 factors:
1, p, …, pα-1, pα
The total number of factors of N equals (α + 1)(β + 1) … (γ + 1)

500 = 2 x 2 x 5 x 5 x 5
So, the number in question should be of the form abq4r4s4 where a, b, q, r, s are primes that minimize abq4r4s4. This is satisfied by 7x11x24x34x54 = 62370000. This marks the end of the First Step which is where we start our search for our magic number.

The next 3 steps would need helper functions defined as below:

As can be seen from the above code, the algorithm to calculate divisors of an integer is as follows:
1. Start by inputting a number n
2. Let an int variable limit = √n
3. Run a loop from i = 1 to  i = limit
    3.1 if n is divisible by i
3.1.1 Add i to the list of divisors
3.1.2 if i and n/i are unequal, add n/i to the list too.
4. End

Finally, executing the 4 steps mentioned earlier can be done like so (the code took less than 2s to arrive at the answer):

Ans: 76576500

Number of Inversions in an Unsorted Array: Python Code

This is my solution to the first programming assignment of Tim Roughgarden’s course on Algorithms  that was due 12:30 PM IST today. Here’s the question quoted as it is:

Programming Question-1
Download the text file here. (Right click and save link as) This file contains all of the 100,000 integers between 1 and 100,000 (inclusive) in some order, with no integer repeated.

Your task is to compute the number of inversions in the file given, where the ith row of the file indicates the ith entry of an array.
Because of the large size of this array, you should implement the fast divide-and-conquer algorithm covered in the video lectures. The numeric answer for the given input file should be typed in the space below.
So if your answer is 1198233847, then just type 1198233847 in the space provided without any space / commas / any other punctuation marks. You can make up to 5 attempts, and we’ll use the best one for grading.
(We do not require you to submit your code, so feel free to use any programming language you want — just type the final numeric answer in the following space.)

My Solution

I modified an earlier code I wrote for merge sort to arrive at the solution. It needed just a couple of modifications, and if you look carefully, it turns out that the number of inversions are unearthed each and every time we merge two sorted sub-arrays. So, intuitively, if the merge sort algorithm was O(nlog2 n), it would take almost as many operations for counting inversions. In python, the code to sort and count inversions in an array of 10,000 integers took less than 3 seconds.

Test my solution here with your own test cases.